转换成0,1模型就好了，注意数据范围

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
#include <assert.h>
using namespace std;
const int MaxM = 500;
const int MaxN = 500;
const int maxnode = MaxM*MaxN;
int K,n,m;
struct DLX
{
	int n,m,size;
	int U[maxnode],D[maxnode],R[maxnode],L[maxnode],Row[maxnode],Col[maxnode];
	int H[MaxN],S[MaxM];
	int ands,ans[MaxN];
	void init(int _n,int _m)
	{
	    ands=0x3f3f3f3f;
		n = _n;
		m = _m;
		for(int i = 0;i <= m;i++)
		{
			S[i] = 0;
			U[i] = D[i] = i;
			L[i] = i-1;
			R[i] = i+1;
		}
		R[m] = 0; L[0] = m;
		size = m;
		for(int i = 1;i <= n;i++)
			H[i] = -1;
	}
	void Link(int r,int c)
	{
		++S[Col[++size]=c];
		Row[size] = r;
		D[size] = D[c];
		U[D[c]] = size;
		U[size] = c;
		D[c] = size;
		if(H[r] < 0)H[r] = L[size] = R[size] = size;
		else
		{
			R[size] = R[H[r]];
			L[R[H[r]]] = size;
			L[size] = H[r];
			R[H[r]] = size;
		}
	}
	void remove(int c)
	{
		for(int i = D[c];i != c;i = D[i])
			L[R[i]] = L[i], R[L[i]] = R[i];
	}
	void resume(int c)
	{
		for(int i = U[c];i != c;i = U[i])
			L[R[i]]=R[L[i]]=i;
	}
	bool v[maxnode];
	int f()
	{
		int ret = 0;
		for(int c = R[0];c != 0;c = R[c])v[c] = true;
		for(int c = R[0];c != 0;c = R[c])
			if(v[c])
			{
				ret++;
				v[c] = false;
				for(int i = D[c];i != c;i = D[i])
					for(int j = R[i];j != i;j = R[j])
						v[Col[j]] = false;
			}
		return ret;
	}
	void Dance(int d)
	{
	    if(d+f()>=ands) return;
		if(R[0] == 0) {ands=min(ands,d);return;}
		int c = R[0];
		for(int i = R[0];i != 0;i = R[i])
			if(S[i] < S[c])
				c = i;
		for(int i = D[c];i != c;i = D[i])
		{
			remove(i);
			for(int j = R[i];j != i;j = R[j])remove(j);
			Dance(d+1);
			for(int j = L[i];j != i;j = L[j])resume(j);
			resume(i);
		}
	}
}g;
int mp[20][20],sa,sb,id[20][20];
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        int tot=0;
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                scanf("%d",&mp[i][j]);
                if(mp[i][j]) id[i][j]=++tot;
            }
        }
        scanf("%d%d",&sa,&sb);
        g.init(n*m,tot);
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                for(int ii=i,cnt=1;cnt<=sa&&ii<n;ii++,cnt++)
                {
                    for(int jj=j,cnt2=1;cnt2<=sb&&jj<m;jj++,cnt2++)
                    {
                        if(mp[ii][jj])
                        {
                            g.Link(i*m+j+1,id[ii][jj]);
                        }
                    }
                }
            }
        }
        g.Dance(0);
        printf("%d\n",g.ands);
    }
    return 0;
}

FZU 1686 神龙的难题 重复覆盖